Problem: $\vec v = (3,-3)$ $3\vec v= ($
Solution: In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $3 \vec{v}$ : $\begin{aligned} {3}\vec v = {3} \cdot (3,-3) &= \left({3} \cdot 3, {3} \cdot (-3)\right) \\\\ &= (9,-9) \end{aligned}$ The answer is $ (9,-9) $.